Ta có:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)

\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)

Đởn giản hết sẽ còn là:

\(\Rightarrow B=\frac{1}{2018}\)

có ai biết câu a, ko vậy

\(A=\frac{1}{2}:\frac{4}{3}:\frac{-5}{4}:\frac{6}{5}:...:\frac{-101}{100}\) 

<=> \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{-4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{-100}{101}\)

Trong biểu thức  A có số số âm là (100-4):2 + 1 =49 số

Vậy A là số âm => \(A=-\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{100}{101}\right)\)

=> \(A=-\left(\frac{1}{2}\cdot\frac{3}{101}\right)=\frac{-3}{202}\)

thanks bn nhiều nha Hiếu

Hì uk nha :D 

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)

\(=\frac{1}{100}\)

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)

Đặt : \(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)

\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)

\(A=\frac{1}{100}\)

Vậy : \(A=\frac{1}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}....\frac{99}{100}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(=\frac{1}{100}\)

Ta có:

 \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

nha

hello